How does a graph with tensile stress-strain curve and compressive stress-strain curve superimposed look like involving true stress and true strain?

Question

If you have a graph with a tensile stress-strain curve of a typical metal or alloy and you superimpose a compressive stress-strain curve on it (from same metal or alloy), how would this look like if you had to sketch it?

I know that when you talking about engineering stress and strain, you will get a typical metal stress-strain curve for tension and then in comparison the compressive one follows same path for the elastic region and then after that rises above the tensile one. And you will notice three differences:

1. compressive one will have no necking and thus no maximum/tensile point.

2. it will fracture differently and with way higher strain

3. like mentioned before; F needed for compression is higher than for tension in the plastic region because of the increasing area and so engineering stress will be higher as well.

Now how would tensile look like versus compression if we talking about real stress and strain? Especially regarding point 3, will compression still rise above tension curve in the plastic region? So in other words will delta F be higher than delta area in compression? Will the extra rise in Force for compression with respect to tension be compensated by the fact that for compression the area increases and for tension it decreases and thus true stress will be equal?

If it not much to ask, could someone sketch a true stress and strain curve with both tension and compression for a typical metal?

Answer 1

It is rare in the mechanical engineering field to have true stress and strain curve diagrams on compression but some of my laboratories were done on it. Here is a diagram of true stress and strain curve in tension, and here is a diagram of true stress and strain curve in compression .

Compression will still rise above tension curve in the plastic region, according to the pictures provided; but you cannot have a general talk about simply "typical metals". Each material has its' own properties and might be completely different.

What you mentioned about the extra rise in force for compression with respect to tension is indeed compensated by the fact that for compression the area increases and for tension it decreases, due to necking. Necking is a type of tensile deformation where relatively large amounts of strain localize disproportionately in a small region of the material, therefore causing the tension to decrease.

Answer 2

One real-life complication is that once you get into the plastic region the behaviour can diverge wildly depending on the geometry and work hardening characteristics of the sample.

For example a cylinder of a fairly ductile material may just get flatter and flatter indefinitely untill you get a flat disk or it exceeds the bounds of the ram compressing it clearly in this case the effective area will increase so true stress will decrease relative to the applied force. This is not dissimilar to some real world forming processes like setting solid rivets.

Alternatively you could get relatively unpredictable failure mechanisms like buckling, mushrooming or rupture at some point especially if work hardening is significant equally if the sample is not dead square relative to the test rig or the sample is not perfectly homogeneous you could start to see bending forces develop.

In summary tension is pretty much just tension regardless of the dimensions of the sample but response to compression is much more dependant on specific test conditions so some care is needed in making direct comparisons.